The Nevanlinna class is a bounded quotient class

(ii) implies (i). Assume $f=g\mathrm{/}h$ with $g,h\in H^{\mathrm{\infty }}(\mathbb{D})$ and $h$ zero-free. Put $\lambda :=\max \{\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }},\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }},1\}$ and replace $(g,h)$ by $(g\mathrm{/}\lambda ,h\mathrm{/}\lambda )$; this leaves $f=g\mathrm{/}h$ unchanged and gives $\mathrm{\mid }g\mathrm{\mid }\le 1$ and $\mathrm{\mid }h\mathrm{\mid }\le 1$ on $\mathbb{D}$. Then $H:={\log }^{+}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}-\log \mathrm{\mid }h\mathrm{\mid }$ is harmonic on $\mathbb{D}$, because $\log \mathrm{\mid }h\mathrm{\mid }$ is harmonic and the first term is constant, and $H\ge 0$ because $\mathrm{\mid }h\mathrm{\mid }\le 1$; moreover, at every point where $\log \mathrm{\mid }f\mathrm{\mid }>0$ one has $\log \mathrm{\mid }f\mathrm{\mid }=\log \mathrm{\mid }g\mathrm{\mid }-\log \mathrm{\mid }h\mathrm{\mid }\le \log \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}-\log \mathrm{\mid }h\mathrm{\mid }\le H$, while where $\log \mathrm{\mid }f\mathrm{\mid }\le 0$ one has ${\log }^{+}\mathrm{\mid }f\mathrm{\mid }=0\le H$. Hence ${\log }^{+}\mathrm{\mid }f\mathrm{\mid }\le H$ with $H$ harmonic on $\mathbb{D}$, so $f\in N(\mathbb{D})$.

(i) implies (ii). Assume $f\in N(\mathbb{D})$ and let $h_0$ be a harmonic majorant of ${\log }^{+}\mathrm{\mid }f\mathrm{\mid }$ on $\mathbb{D}$; then $h_0\ge {\log }^{+}\mathrm{\mid }f\mathrm{\mid }\ge 0$ on $\mathbb{D}$. By [L3] there is a harmonic conjugate ${\tilde{h}}_0$ of $h_0$ on $\mathbb{D}$, and $h:=\exp (-h_0-i{\tilde{h}}_0)$ is holomorphic, zero-free and satisfies $\mathrm{\mid }h\mathrm{\mid }=e^{-h_0}\le 1$ on $\mathbb{D}$; hence $h\in H^{\mathrm{\infty }}(\mathbb{D})$ with $\mathrm{\mid }h\mathrm{\mid }\le 1$.

The companion numerator. With $h$ as in step 1.2 put $g:=fh$. Then $g$ is holomorphic on $\mathbb{D}$, and for every $z$ with $f(z)\mathrm{\ne }0$, $\mathrm{\mid }g(z)\mathrm{\mid }=\mathrm{\mid }f(z)\mathrm{\mid }{e}^{-h_0(z)}=e^{\log \mathrm{\mid }f(z)\mathrm{\mid }-h_0(z)}\le e^{{\log }^{+}\mathrm{\mid }f(z)\mathrm{\mid }-h_0(z)}\le 1,$ while $\mathrm{\mid }g(z)\mathrm{\mid }=0\le 1$ at the zeros of $f$; here we used $\log \mathrm{\mid }f\mathrm{\mid }\le {\log }^{+}\mathrm{\mid }f\mathrm{\mid }\le h_0$. Thus $g\in H^{\mathrm{\infty }}(\mathbb{D})$ with $\mathrm{\mid }g\mathrm{\mid }\le 1$, and $f=g\mathrm{/}h$ because $h$ is zero-free. This proves (i)$\Rightarrow$(ii) with both functions bounded by $1$.

Assembly and non-uniqueness. Step 1.1 proves (ii)$\Rightarrow$(i) and steps 1.2 and 2.1 prove (i)$\Rightarrow$(ii), with the normalization $\mathrm{\mid }g\mathrm{\mid }\le 1$, $\mathrm{\mid }h\mathrm{\mid }\le 1$ established in each direction; hence (i) and (ii) are equivalent. If $f=g\mathrm{/}h$ with $h$ zero-free and $\phi \in H^{\mathrm{\infty }}$ is zero-free, then $g\phi ,h\phi \in H^{\mathrm{\infty }}$, $h\phi$ is zero-free and $(g\phi )\mathrm{/}(h\phi )=g\mathrm{/}h=f$, so the representation is not unique. The construction used only the harmonic conjugate and the exponential, both choice-free, so neither the Axiom of Choice nor countable choice is used.